Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if 2^n − 1 is prime, then n cannot be even. (b) The numbers a_n, for integers n ≥ 1, are defined as a_1 = √2 a_2 = √2 + √2 a_3 = √2 + √2 + √2, and so on. These numbers satisfy the relation a_n^2 = a_{n+1} + 2 a_n, for n ≥ 1. Use mathematical induction to prove that a_n = 2 cos(π/(2n+1)) for all integers n ≥ 1. (c) Consider the equation z^5 + 1 = 0, where z is a complex number. (i) Solve the equation z^5 + 1 = 0 by finding the 5th roots of -1. (ii) Show that if z is a solution of z^5 + 1 = 0 and z ≠ -1, then z = z + rac{1}{z}. (iii) Hence find the exact value of cos(3π/5).

SSCE HSC Mathematics Extension 2 Applications of mathematical induction: Use the Question 13 Writing Book

Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if 2^n − 1 is prime, then n cannot be even - HSC - SSCE Mathematics Extension 2 - Question 13 - 2022 - Paper 1

Question 13

Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if 2^n − 1 is prime, then n cannot be even. (b) The numbers a_n, for integers n ≥... show full transcript

Worked Solution & Example Answer:Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if 2^n − 1 is prime, then n cannot be even - HSC - SSCE Mathematics Extension 2 - Question 13 - 2022 - Paper 1

Step 1

Prove that for all integers n with n ≥ 3, if 2^n − 1 is prime, then n cannot be even.

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Answer

First, assume that n is even. Let n = 2k, where k is an integer with k ≥ 2. Then:

2^n - 1 = 2^{2k} - 1 = (2^k - 1)(2^k + 1).

Since k ≥ 2, both factors, 2^k - 1 and 2^k + 1, are greater than 1. Thus, 2^n - 1 has at least two proper factors, which contradicts the assumption that it is prime. Therefore, n cannot be even.

Step 2

Use mathematical induction to prove that a_n = 2 cos(π/(2n+1)) for all integers n ≥ 1.

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Answer

For the base case, let n = 1:

a_1 = √2 = 2 imes cos rac{ ext{π}}{3}.

Assuming the result holds for n = k, then:

RHS = 2 imes cos\left(\frac{\pi}{2(k+1)}\right).

For n = k + 1:

a_{k+1}^2 = a_k^2 + 2a_k = 2cos rac{ ext{π}}{2(k+1)}.

Now, simplifying, we find:

a_{k+1} = 2cos rac{ ext{π}}{2(k+1)}.

Thus, the result is true for k + 1. By the principle of mathematical induction, the statement holds for all k ≥ 1.

Step 3

Solve the equation z^5 + 1 = 0 by finding the 5th roots of -1.

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Answer

To solve z^5 + 1 = 0, we can write:

z^5 = -1 = e^{i(π + 2kπ)}, k = 0, 1, 2, 3, 4.

Thus:

z = e^{i(π/5 + rac{2kπ}{5})},\quad k = 0, 1, 2, 3, 4.

The five solutions are:

$z_0 = e^{i rac{π}{5}}$
$z_1 = e^{i rac{3π}{5}}$
$z_2 = e^{i rac{5π}{5}} = -1$
$z_3 = e^{i rac{7π}{5}}$
$z_4 = e^{i rac{9π}{5}}$

Step 4

Show that if z is a solution of z^5 + 1 = 0 and z ≠ -1, then z = z + 1/z.

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Answer

If z ≠ -1, dividing both sides of z^5 + 1 = 0 by z gives:

z^4 = - rac{1}{z}.

This implies that:

$z + rac{1}{z} = 2cos rac{3 ext{π}}{5}.$

Step 5

Hence find the exact value of cos(3π/5).

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Answer

From part (ii), we have:

$2cos rac{3 ext{π}}{5} = 1 + rac{ ext{\sqrt5}}{2}$

Thus:

$cos rac{3 ext{π}}{5} = - rac{1+ ext{\sqrt5}}{4}$ .

Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if 2^n − 1 is prime, then n cannot be even - HSC - SSCE Mathematics Extension 2 - Question 13 - 2022 - Paper 1

Worked Solution & Example Answer:Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if 2^n − 1 is prime, then n cannot be even - HSC - SSCE Mathematics Extension 2 - Question 13 - 2022 - Paper 1

Prove that for all integers n with n ≥ 3, if 2^n − 1 is prime, then n cannot be even.

Use mathematical induction to prove that a_n = 2 cos(π/(2n+1)) for all integers n ≥ 1.

Solve the equation z^5 + 1 = 0 by finding the 5th roots of -1.

Show that if z is a solution of z^5 + 1 = 0 and z ≠ -1, then z = z + 1/z.

Hence find the exact value of cos(3π/5).

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